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CAT 2020 Quant Question [Slot 3] with Solution 11

Question

Let m and n be positive integers, If \({x^2} + mx + 2n = 0\) and \({x^2} + 2nx + m = 0\) have real roots, then the smallest possible value of \(m + n\) is

  1. 7
  2. 8
  3. 5
  4. 6
Option: 4
Solution:

Since the roots are real \({m^2} - 8n \ge 0\) and \({(2n)^2} - 4m \ge 0\) \( \Rightarrow {n^2} - m \ge 0\)

\( \Rightarrow {n^4} \ge {m^2} \ge 8n\)

\( \Rightarrow n \ge 2\) and \(m \ge 4\)

Hence the least value of \(m + n = 2 + 4 = 6\)

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